Find the value of alpha on a pentagon.

Question

Here is the question:

I have been trying this out for an hour, and I still dont know how to do it. Here's my progress: I'm stuck in this exact position and I dont know what to do next.

Can anyone help?

Answer 1

$$\triangle BCG\implies\frac{BC}{\sin 45^\circ}=\frac{CG}{\sin63^\circ}\implies CG=a\frac{\sin63^\circ}{\sin45^\circ}$$

$$CG=a\frac{\sin45^\circ\cos18^\circ+\cos45^\circ\sin18^\circ}{\sin45^\circ}=a(\cos18^\circ+\sin18^\circ)\tag{1}$$

$$EC=2a\cos36^\circ=2a(\cos^2 18^\circ-\sin^2 18^\circ)\tag{2}$$

$$ (1) \land (2) \implies EG=EC-CG=a(2\cos^2 18^\circ-2\sin^2 18^\circ-\cos18^\circ-\sin18^\circ)\tag{3}$$

$$\triangle EGH\ (\angle EGH=\alpha) \implies \frac{EH}{\sin\alpha}=\frac{a}{2\sin\alpha}=\frac{EG}{\sin(180^\circ-36^\circ-\alpha)}=\frac{EG}{\sin(36^\circ+\alpha)}$$

$$\frac{a \sin(36^\circ+\alpha)}{2\sin\alpha}=EG=a(2\cos^2 18^\circ-2\sin^2 18^\circ-\cos18^\circ+\sin18^\circ)$$

$$\frac{\sin36^\circ\cos\alpha+\cos36^\circ\sin\alpha}{2\sin\alpha}=2\cos^2 18^\circ-2\sin^2 18^\circ-\cos18^\circ-\sin18^\circ$$

$$\sin18^\circ\cos18^\circ\cot\alpha+\frac12(\cos^2 18^\circ-\sin^2 18^\circ)=2\cos^2 18^\circ-2\sin^2 18^\circ-\cos18^\circ-\sin18^\circ$$

$$\cot\alpha=\frac{\frac32(\cos^2 18^\circ-\sin^2 18^\circ)-\cos18^\circ-\sin18^\circ}{\sin18^\circ\cos18^\circ}\tag{4}$$

$$\cot\alpha=\frac{3\cos36^\circ}{\sin36^\circ}-\frac{\cos18^\circ+\sin18^\circ}{\sin18^\circ\cos18^\circ}$$

$$\cot\alpha=3\cot36^\circ-\sec18^\circ-\csc18^\circ$$

For oblique angle $\alpha$ this gives the result:

$$\alpha=99^\circ$$

EDIT Let us prove that the answer is indeed $99^\circ$ and not some angle very close to it.

Introduce a constant:

$$t=\tan9^\circ$$

Obviously:

$$\tan(5\times 9^\circ)-1=0\tag{5}$$

This can be expanded so that the only items figuring in (5) are $t=\tan9^\circ$. After some straightforward but difficult work this leads to:

$$(-1 + t) (1 - 4 t - 14 t^2 - 4 t^3 + t^4) = 0$$

or (beacuse $t\ne1)$

$$1 - 4 t - 14 t^2 - 4 t^3 + t^4 = 0\tag{6}$$

Now, introduce the following replacements:

$$\cos18^\circ=\frac{1-t^2}{1+t^2}\ \ \sin18^\circ=\frac{2t}{1+t^2}$$

...and replace it into (4). You get:

$$\cot\alpha=\frac{1-4t - 18t^2 -4 t^3 + 5 t^4}{4 t-4 t^3}$$

$$\cot\alpha=\frac{(1-4t - 14t^2 -4 t^3 + t^4)+ (-4t^2+4t^4)}{4 t-4 t^3}$$

The first item in the numerator is equal to zero according to (6) so the final result is:

$\cot\alpha=-t=-\tan 9^\circ$ which is only possible for $\alpha=99^\circ$, as stated initially.

Answer 2

Let us put letters on the pentagon, let it be $ABCDE$, and let $F$ be the mid point of $DE$ as in the picture:

Let $I$ be the incenter of $\Delta BEF$, which has the angles with measures $90^\circ=2\cdot 45^\circ$ in $F$, $72^\circ=2\cdot 36^\circ$ in $E$, and $18^\circ=2\cdot 9^\circ$ in $B$.

Then the angle $\widehat{BIC}$ is exterior for $\Delta EIB$, it measures $\hat E+\hat B=36^\circ+9^\circ=45^\circ$, so $I\in CD$ constructed above uniquely matches the point from the problem. So in $\Delta EIF$ $$ \widehat{EIF} = 180^\circ - 36^\circ - 45^\circ = 99^\circ\ . $$

$\square$